A defense contractor doubles its production every year, i.e., it produces "1000" units on the first year, "2000" units on the second year, "4000" units on the third year, and so on. After 10 years, how many units will it have produced?
Consider the following example.
A defense contractor doubles its production every year, i.e., it
produces
"1000" units on the first year, "2000" units on the second year, "4000"
units on the third year, and so on. After 10 years, how
many units will it have produced?
The sequence of number of items produced in each year is called a Geometric Series and the sum of this series can be written as follows.
S = 1000 + 1000(21) + 1000(22) + 1000(23) + ... + 1000(29)
If the number of years( in this case, it is 10 ) is very large, then it is
very difficult to find the above sum without using a simple formula.
Let us consider the general case of a "Geometric Series".
Let S be the sum of the first n terms in the series:
The sum S in Geometric Series can be found as follows.
By multiplying the above equation on both sides by r, we get
r.S = a(r1) + a(r2) + a(r3) +... + a(rn-1) + a(rn). Now subtracting this equation from the above equation, we get
Therefore,
where a = the first term and
r = fixed rate that multiplies the
preceding term.
So, we can apply this sum formula to the above example by substituting
a =1000
r = 2
n =10
and get the answer as follows.
Sn= 1000(1-210)/(1-2) = 1000(210-1)=1,023,000.
Did you notice how simple it is to find the sum in the above example using the formula?